java - 如何在没有迭代器的情况下迭代 Set/HashSet？

Question

如何在没有以下内容的情况下迭代Set/ HashSet？

Iterator iter = set.iterator();
while (iter.hasNext()) {
    System.out.println(iter.next());
}

Answer 1

您可以使用增强的 for 循环：

Set<String> set = new HashSet<String>();

//populate set

for (String s : set) {
    System.out.println(s);
}

或者使用 Java 8：

set.forEach(System.out::println);

Answer 2

至少有六种额外的方法可以迭代一个集合。以下是我所知道的：

方法一

// Obsolete Collection
Enumeration e = new Vector(movies).elements();
while (e.hasMoreElements()) {
  System.out.println(e.nextElement());
}

方法二

for (String movie : movies) {
  System.out.println(movie);
}

方法三

String[] movieArray = movies.toArray(new String[movies.size()]);
for (int i = 0; i < movieArray.length; i++) {
  System.out.println(movieArray[i]);
}

方法四

// Supported in Java 8 and above
movies.stream().forEach((movie) -> {
  System.out.println(movie);
});

方法五

// Supported in Java 8 and above
movies.stream().forEach(movie -> System.out.println(movie));

方法6

// Supported in Java 8 and above
movies.stream().forEach(System.out::println);

这是HashSet我用于示例的：

Set<String> movies = new HashSet<>();
movies.add("Avatar");
movies.add("The Lord of the Rings");
movies.add("Titanic");

Answer 3

将您的集合转换为数组 也可以帮助您迭代元素：

Object[] array = set.toArray();

for(int i=0; i<array.length; i++)
   Object o = array[i];

Answer 4

为了演示，请考虑以下集合，其中包含不同的 Person 对象：

Set<Person> people = new HashSet<Person>();
people.add(new Person("Tharindu", 10));
people.add(new Person("Martin", 20));
people.add(new Person("Fowler", 30));

人物模型类

public class Person {
    private String name;
    private int age;

    public Person(String name, int age) {
        this.name = name;
        this.age = age;
    }

    //TODO - getters,setters ,overridden toString & compareTo methods

}

1. for 语句有一种设计用于通过集合和数组进行迭代的形式。这种形式有时被称为增强的 for 语句，可用于使您的循环更加紧凑和易于阅读。

for(Person p:people){
  System.out.println(p.getName());
}

2. Java 8 - java.lang.Iterable.forEach（消费者）

people.forEach(p -> System.out.println(p.getName()));

default void forEach(Consumer<? super T> action)

Performs the given action for each element of the Iterable until all elements have been processed or the action throws an exception. Unless otherwise specified by the implementing class, actions are performed in the order of iteration (if an iteration order is specified). Exceptions thrown by the action are relayed to the caller. Implementation Requirements:

The default implementation behaves as if: 

for (T t : this)
     action.accept(t);

Parameters: action - The action to be performed for each element

Throws: NullPointerException - if the specified action is null

Since: 1.8

Answer 5

您可以使用函数式操作来获得更简洁的代码

Set<String> set = new HashSet<String>();

set.forEach((s) -> {
     System.out.println(s);
});

Answer 6

以下是一些关于如何迭代 Set 及其性能的提示：

public class IterateSet {

    public static void main(String[] args) {

        //example Set
        Set<String> set = new HashSet<>();

        set.add("Jack");
        set.add("John");
        set.add("Joe");
        set.add("Josh");

        long startTime = System.nanoTime();
        long endTime = System.nanoTime();

        //using iterator
        System.out.println("Using Iterator");
        startTime = System.nanoTime();
        Iterator<String> setIterator = set.iterator();
        while(setIterator.hasNext()){
            System.out.println(setIterator.next());
        }
        endTime = System.nanoTime();
        long durationIterator = (endTime - startTime);


        //using lambda
        System.out.println("Using Lambda");
        startTime = System.nanoTime();
        set.forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationLambda = (endTime - startTime);


        //using Stream API
        System.out.println("Using Stream API");
        startTime = System.nanoTime();
        set.stream().forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationStreamAPI = (endTime - startTime);


        //using Split Iterator (not recommended)
        System.out.println("Using Split Iterator");
        startTime = System.nanoTime();
        Spliterator<String> splitIterator = set.spliterator();
        splitIterator.forEachRemaining((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationSplitIterator = (endTime - startTime);


        //time calculations
        System.out.println("Iterator Duration:" + durationIterator);
        System.out.println("Lamda Duration:" + durationLambda);
        System.out.println("Stream API:" + durationStreamAPI);
        System.out.println("Split Iterator:"+ durationSplitIterator);
    }
}

该代码是不言自明的。

持续时间的结果是：

Iterator Duration: 495287
Lambda Duration: 50207470
Stream Api:       2427392
Split Iterator:    567294

我们可以看到Lambda耗时最长而Iterator最快。

Answer 7

枚举（？）：

Enumeration e = new Vector(set).elements();
while (e.hasMoreElements())
    {
        System.out.println(e.nextElement());
    }

另一种方式（java.util.Collections.enumeration()）：

for (Enumeration e1 = Collections.enumeration(set); e1.hasMoreElements();)
    {
        System.out.println(e1.nextElement());
    }

爪哇 8：

set.forEach(element -> System.out.println(element));

或者

set.stream().forEach((elem) -> {
    System.out.println(elem);
});

Answer 8

但是，已经有非常好的答案可用。这是我的答案：

1. set.stream().forEach(System.out::println); // It simply uses stream to display set values
2. set.forEach(System.out::println); // It uses Enhanced forEach to display set values

此外，如果此 Set 是自定义类类型，例如：客户。

Set<Customer> setCust = new HashSet<>();
    Customer c1 = new Customer(1, "Hena", 20);
    Customer c2 = new Customer(2, "Meena", 24);
    Customer c3 = new Customer(3, "Rahul", 30);

setCust.add(c1);
setCust.add(c2);
setCust.add(c3);
    setCust.forEach((k) -> System.out.println(k.getId()+" "+k.getName()+" "+k.getAge()));

// 客户类：

class Customer{
private int id;
private String name;
private int age;

public Customer(int id,String name,int age){
this.id=id;
this.name=name;
this.age=age;
} // Getter, Setter methods are present.}